Let f : R to R be defined by f(x) = frac{|x| | vedclass

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(D) Given $f(x) = \frac{|x| - 1}{|x| + 1}$.For a function to be one-one,if $f(x_1) = f(x_2)$,then $x_1$ must be equal to $x_2$.Let $f(x_1) = f(x_2)$:$

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neither one-one nor onto

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(D) Given $f(x) = \frac{|x| - 1}{|x| + 1}$.For a function to be one-one,if $f(x_1) = f(x_2)$,then $x_1$ must be equal to $x_2$.Let $f(x_1) = f(x_2)$:$\frac{|x_1| - 1}{|x_1| + 1} = \frac{|x_2| - 1}{|x_2| + 1}$$(|x_1| - 1)(|x_2| + 1) = (|x_2| - 1)(|x_1| + 1)$$|x_1||x_2| + |x_1| - |x_2| - 1 = |x_1||x_2| - |x_1| + |x_2| - 1$$2|x_1| = 2|x_2| \implies |x_1| = |x_2|$This implies $x_1 = x_2$ or $x_1 = -x_2$. Since $f(1) = f(-1) = 0$,the function is not one-one (it is many-one).For onto: Let $y = \frac{|x| - 1}{|x| + 1}$.$y(|x| + 1) = |x| - 1 \implies y|x| + y = |x| - 1 \implies |x|(y - 1) = -1 - y \implies |x| = \frac{1 + y}{1 - y}$.Since $|x| \ge 0$,we must have $\frac{1 + y}{1 - y} \ge 0$. Solving this inequality,we get $y \in [-1, 1)$.The range of $f$ is $[-1, 1)$,which is not equal to the codomain $R$. Therefore,the function is not onto.

Let $f : R \to R$ be defined by $f(x) = \frac{|x| - 1}{|x| + 1}$. Then $f$ is

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